Question

find value of a and b

Answer 1

◀️hey mate! here's your answer!▶️
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$\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} }$
then, we will rationalize them,
$\\ = \frac{(7 + 3 \sqrt{5})(3 - \sqrt{5} ) }{(3 + \sqrt{5} )(3 - \sqrt{5} )} - \frac{(7 - 3 \sqrt{5})(3 + \sqrt{5} ) }{(3 - \sqrt{5} )(3 + \sqrt{5} )}$

we know that, (a+b)(a-b) = a² - b² , so using this identity in the denominator,
$\\ = \frac{7(3 - \sqrt{5} ) + 3 \sqrt{5}(3 - \sqrt{5}) }{ {(3)}^{2} - { (\sqrt{5}) }^{2} } - \frac{7(3 + \sqrt{5}) - 3 \sqrt{5}(3 + \sqrt{5})}{ {(3) ^{2} - { (\sqrt{5}) }^{2} }}$
$\\ = \frac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 }{9 - 5} - \frac{21 + 7 \sqrt{5} - 9 \sqrt{5} + 15}{9 - 5}$
$\\ = \frac{6 + 2 \sqrt{5} }{4} - \frac{6 - 2 \sqrt{5} }{4}$

now, we will take LCM,
$\\ = \frac{6 + 2 \sqrt{5 } - (6 - 2 \sqrt{5}) }{4}$
$\\ = \frac{6 + 2 \sqrt{5} - 6 + 2 \sqrt{5} }{4}$
$\\ = \frac{6 - 6 + 4 \sqrt{5} }{4}$
$\\ = \frac{4 \sqrt{5} }{4}$

so, here,

$a = \frac{1}{4} \\ \\ and \\ \\ b \sqrt{5} = \frac{4 \sqrt{5} }{4}$

here, √5 cancels from both side!
$\\ = >b = \frac{4}{4} \\ \\ = > b = 1$

hope you understood!